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BT-203 · Basic Mechanical Engineering/Quick Revision Short Notes

Basic Mechanical Engineering (BT-203) - Unit 1 Short Notes

How unit 1 is examined

This unit covers classification of materials, cast iron and steels, alloy steels, mechanical properties, the tensile test, Hooke's law and hardness testing; the tensile test, Hooke's law numericals and the classification and hardness explanations carry most marks.

Classification of engineering materials

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Medium weight</span>

Definition. Engineering materials are the substances used to make machines and structures; they are grouped by composition into metals, ceramics, polymers and composites. An alloy is a homogeneous mixture of two or more metals, or of a metal with a non-metal, having metallic properties.

Diagram.

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Key points.

  1. Ferrous metals have iron as the base (cast iron, carbon steel, alloy steel) and are strong and cheap, used for shafts, gears, frames and machine beds.
  2. Non-ferrous metals have no iron base (copper, aluminium, zinc, lead, tin) and are light, corrosion resistant and conductive, used for wires, utensils and aircraft parts.
  3. Ceramics are inorganic non-metallic solids (glass, porcelain, alumina, brick) that are hard and heat resistant but brittle, used for insulators, refractories and cutting tools.
  4. Polymers are long-chain organic materials: thermoplastics soften on heating (PVC, nylon), thermosets set permanently (bakelite), and elastomers stretch elastically (rubber); they are used for pipes, gears, tyres and casings.
  5. Composites combine two materials to get properties neither has alone (glass fibre in resin, reinforced concrete), used for boat hulls, sports goods and aircraft panels.
  6. Alloying improves strength, hardness, corrosion resistance and wear resistance compared with the pure metal.
  7. Alloy applications: steel (Fe + C) for structures and tools, brass (Cu + Zn) for valves and fittings, bronze (Cu + Sn) for bearings and gears, duralumin (Al + Cu) for aircraft.

<mark>An alloy is a homogeneous mixture of two or more metals, or of a metal and a non-metal, with properties better than those of its components.</mark>

Answer frame. Open with the definition of engineering materials; draw the classification tree; develop metals (ferrous, non-ferrous), ceramics, polymers, composites with one application each; then define alloy and list reasons and applications; close with one line on why alloys are preferred.

Asked: [7 marks] (Nov 2022, Dec 2024, Jun 2025) Classify engineering materials. What do you mean by alloys? Write some applications of alloys. Explain in detail classification of engineering material along their applications.

Composition of cast iron and carbon steels

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Low weight</span>

Definition. Cast iron is an iron-carbon alloy with 2-4.5% carbon (plus Si, Mn, S, P); carbon steel is an iron-carbon alloy with up to about 1.5% carbon.

Key points.

  1. Grey cast iron has carbon as graphite flakes, is easy to cast and machine and damps vibration; it is used for machine beds, engine blocks and pipes.
  2. White cast iron has carbon as hard iron carbide (cementite), so it is very hard, wear resistant and brittle; it is used for grinding balls and liners.
  3. Malleable cast iron is white iron heat treated to give nodules of carbon, so it is tougher and ductile; it is used for pipe fittings, brackets and automobile parts.
  4. Ductile (nodular, SG) cast iron has graphite as spheres, giving high strength and toughness; it is used for crankshafts, gears and pressure pipes.
  5. Carbon steels: low or mild (up to 0.25% C) is ductile and weldable, medium (0.25-0.6% C) is stronger, high (0.6-1.5% C) is hard and used for tools, springs and dies.

Asked: [7 marks] (Dec 2024) Write in detail classification of cast irons along their applications.

Iron carbon diagram

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Not asked since 2022</span>

Definition. The iron-carbon (Fe-Fe3C) diagram is a phase diagram plotting temperature against carbon percentage, showing the phases of steel and cast iron at equilibrium.

Key points.

  1. Ferrite is nearly pure iron (BCC, soft and ductile) and austenite is the FCC solid solution of carbon in iron, non-magnetic and stable above 723 C.
  2. Cementite (Fe3C) is the hard, brittle carbide containing 6.67% carbon.
  3. The eutectoid point is at 0.8% C and 723 C, where austenite changes to pearlite (ferrite + cementite); the eutectic point is at 4.3% C and 1147 C.
  4. Steels lie below 2% carbon and cast irons above 2%.

Alloy steels and their applications

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Medium weight</span>

Definition. Alloy steel is steel to which elements such as Cr, Ni, Mo, W, V, Mn or Si are added deliberately, beyond the small amounts in plain carbon steel, to improve its properties.

Key points.

  1. Chromium increases hardness, wear resistance and corrosion resistance; above about 12% it gives stainless steel.
  2. Nickel increases toughness, strength and corrosion resistance without losing ductility.
  3. Molybdenum increases strength at high temperature, hardenability and creep resistance, and reduces temper brittleness.
  4. Tungsten keeps hardness at high temperature ("red hardness"), so it is used in high-speed tool steel.
  5. Vanadium refines the grain and increases strength, toughness and fatigue resistance.
  6. Manganese increases strength, hardness and wear resistance; about 13% Mn gives a work-hardening, abrasion-resistant steel.
  7. Silicon increases strength and elasticity (spring steel) and gives good magnetic properties for transformer cores.
  8. Cobalt increases hot hardness and magnetic strength; titanium prevents intergranular corrosion in stainless steel.
Point Plain carbon steel Alloy steel
Composition Iron + carbon, little Mn and Si Iron + carbon + deliberate alloying elements
Hardenability Low, only thin sections harden High, deep hardening
Strength and toughness Moderate Higher
Corrosion resistance Poor Good (Cr, Ni)
Hot hardness Lost early Retained (W, Mo)
Cost Low Higher

Necessity of alloying: pure metals and plain steel lack the strength, hardness, corrosion and heat resistance needed in modern machines. Applications: stainless steel for utensils, chemical plant and surgical tools; Ni-Cr steel for gears and shafts; high-speed steel for cutting tools; Si-Mn steel for springs.

<mark>Alloying elements are added to steel to raise strength, hardness, toughness, hardenability and corrosion resistance beyond what plain carbon steel can give.</mark>

Answer frame. For the 8-mark question, open with the definition of alloy steel, then give eight elements in a two-column table of element and effect, and close with applications. For the 7-mark question, open with the necessity of alloying, define alloy steel, then draw the comparison table; close with one application.

Asked: [8 marks] (Jun 2023) Explain the effect of any 8 alloying elements in alloy steel with details. Asked: [7 marks] (Dec 2024) What is the necessity of alloying? What are alloy steels, and how do they differ from carbon steels?

Mechanical properties like strength, hardness, toughness, ductility, brittleness, malleability etc. of materials

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Medium weight</span>

Definition. Mechanical properties describe how a material responds to applied loads.

Key points.

  1. Strength is the ability to resist deformation or failure under load without breaking; it may be tensile, compressive, shear, yield or ultimate strength (example: steel).
  2. Elasticity is the ability to regain the original shape after the load is removed (example: spring steel).
  3. Plasticity is the ability to undergo permanent deformation without fracture (example: hot clay, lead).
  4. Ductility is the ability to be drawn into wires under tensile load, measured by percentage elongation and reduction in area (example: copper, mild steel).
  5. Malleability is the ability to be hammered or rolled into thin sheets without cracking (example: gold, aluminium).
  6. Hardness is resistance to indentation, scratching or abrasion (example: hardened tool steel, diamond).
  7. Toughness is the ability to absorb energy up to fracture, that is resistance to impact (example: mild steel, wrought iron).
  8. Brittleness is failure with little or no plastic deformation (example: cast iron, glass). Also fatigue (failure under repeated loading) and creep (slow deformation under constant load at high temperature).
  9. Tensile strength is the maximum stress a material bears in tension before breaking.

<mark>Ductility is the property of a material to undergo large permanent deformation in tension, such as being drawn into wire, before it fractures.</mark>

Answer frame. Open by listing the properties (strength, elasticity, plasticity, ductility, malleability, hardness, toughness, brittleness, fatigue, creep); define the three or four asked, each with a one-line example; close by contrasting ductile and brittle materials.

Asked: [7 marks] (Jun 2022, Dec 2023, Jun 2025) Enlist important mechanical properties of engineering materials. Define any three with suitable examples. Define strength and ductility. Define toughness, brittleness, tensile strength and ductility.

Tensile test: stress-strain diagram of ductile and brittle materials

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>

Definition. The tensile test pulls a standard specimen to fracture on a Universal Testing Machine (UTM) while load and elongation are recorded, giving the stress-strain diagram from which strength and ductility are found.

Experimental set up and procedure.

  1. The UTM has a fixed frame, a fixed upper crosshead, a movable lower crosshead, grips holding the specimen, a load cell or dial to read load, and an extensometer to measure elongation.
  2. A round mild steel specimen is machined with a gauge length $L_0$ (usually $5.65\sqrt{A_0}$) and diameter $d_0$ is measured.
  3. The specimen is gripped, the extensometer fitted, and the load applied slowly and steadily.
  4. Load and elongation are noted at regular steps until fracture; stress $\sigma = P/A_0$ and strain $\epsilon = \Delta L/L_0$ are plotted.
  5. After fracture the final length $L_f$ and neck diameter $d_f$ are measured.

Diagram.

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Key points.

  1. From O to A stress is proportional to strain, so Hooke's law holds; A is the proportional limit and the slope is Young's modulus.
  2. Up to B, the elastic limit, the specimen returns to its original length on unloading.
  3. At C (upper yield) the stress drops to D (lower yield) and the specimen stretches at nearly constant stress; permanent (plastic) deformation begins.
  4. After yielding, strain hardening raises the stress up to E, the ultimate stress, the maximum nominal stress.
  5. Beyond E the cross-section reduces locally (necking), the load falls and the specimen breaks at F.
  6. Mild steel is ductile: a distinct yield point, large plastic strain and a cup-and-cone fracture.
Point Ductile (mild steel) Brittle (cast iron)
Yield point Clear yield point None
Plastic region Long Almost none
Necking Yes No
Elongation Large (about 20% or more) Very small
Fracture Cup and cone, at lower stress than ultimate Sudden, flat, at the ultimate stress
Energy absorbed High (large area under curve) Low

Formula. $\text{\% elongation}=\frac{L_f-L_0}{L_0}\times100$, $\text{\% reduction in area}=\frac{A_0-A_f}{A_0}\times100$.

Example (Dec 2023). Given $d_0=10$ mm, $L_0=50$ mm, $P_u=60$ kN, $P_b=40$ kN, $L_f=55$ mm, $d_f=8$ mm.

Quantity Working Result
$A_0$ $\frac{\pi}{4}(10)^2$ 78.54 mm²
$A_f$ $\frac{\pi}{4}(8)^2$ 50.27 mm²
Ultimate stress $60\times10^3/78.54$ 763.94 N/mm²
Breaking stress $40\times10^3/78.54$ 509.30 N/mm²
True breaking stress $40\times10^3/50.27$ 795.77 N/mm²
% elongation $(55-50)/50\times100$ 10%

Ultimate stress = 763.94 MPa, breaking stress = 509.30 MPa, true breaking stress = 795.77 MPa, elongation = 10%.

Definitions (6 marks). Young's modulus $E=\sigma/\epsilon$ within the proportional limit, unit N/m² (Pa). Elastic limit is the maximum stress up to which the material returns to its original shape on unloading. Ultimate point is the maximum nominal stress the material sustains before necking begins.

<mark>The stress-strain diagram of mild steel shows a proportional limit, elastic limit, upper and lower yield points, ultimate stress and a breaking point, whereas a brittle material breaks suddenly without yielding or necking.</mark>

Answer frame. For the set-up question, open with the purpose of the tensile test; draw the UTM sketch and name its parts; give the procedure, then draw the mild steel curve with points A to F and the brittle curve, and close with the comparison table. For the numerical, write Given, areas, each stress, then the bold answers. For the definitions, one sentence each.

Asked: [7 marks] (Jun 2022, Jun 2023, Jun 2025) Explain the experimental set up of tensile testing of steel. Also compare stress-strain diagrams of ductile and brittle materials. Explain the stress-strain diagram for ductile materials with neat sketch. Describe the procedure of a tensile test. Asked: [8 marks] (Dec 2023) Mild steel specimen, d = 10 mm, gauge length 50 mm, ultimate load 60 kN, breaking load 40 kN, length at rupture 55 mm, neck diameter 8 mm: find ultimate stress, breaking stress, true breaking stress, percentage elongation. Asked: [6 marks] (Jun 2023) Define Young's modulus, elastic limit and ultimate point. Pitfall: True breaking stress uses the neck area $A_f$, nominal breaking stress uses the original area $A_0$; mixing them loses marks.

Hooke's law and modulus of elasticity

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>

Definition. Hooke's law states that within the elastic (proportional) limit, stress is directly proportional to strain. The constant of proportionality is the modulus of elasticity.

Formula. $$\sigma=E\epsilon,\qquad \sigma=\frac{P}{A},\qquad \epsilon=\frac{\Delta L}{L},\qquad \Delta L=\frac{PL}{AE}$$

Key points.

  1. Young's modulus $E=\sigma/\epsilon$ is for tension or compression, in N/mm² or GPa (steel about 200 GPa, brass about 90 GPa).
  2. The modulus of rigidity $G=\tau/\gamma$ is shear stress over shear strain.
  3. The bulk modulus $K=p/(\Delta V/V)$ is volumetric stress over volumetric strain.
  4. Strain has no unit, being a ratio of lengths.
  5. A larger $E$ means a stiffer material; the modulus is the slope of the straight part of the stress-strain curve.

Example 1 (brass rod, Nov 2022, Jun 2022). $d=25$ mm, $L=250$ mm, $P=50$ kN, $\Delta L=0.3$ mm.

$A=\frac{\pi}{4}(25)^2=490.87$ mm²; $E=\frac{PL}{A\Delta L}=\frac{50\times10^3\times250}{490.87\times0.3}$ = 84,883 N/mm².

E = 84.88 GPa (about 84.9 GPa).

Example 2 (steel rod). $L=200$ cm $=2000$ mm, $d=30$ mm, $P=30$ kN, $E=2\times10^5$ N/mm².

$A=\frac{\pi}{4}(30)^2=706.86$ mm²; $\sigma=30000/706.86=42.44$ N/mm²; $\epsilon=42.44/(2\times10^5)=2.122\times10^{-4}$; $\Delta L=\epsilon L=2.122\times10^{-4}\times2000=0.424$ mm.

Stress = 42.44 N/mm², strain = 2.122 x 10^-4, elongation = 0.424 mm.

==Hooke's law: within the elastic limit, stress is directly proportional to strain, so $\sigma = E\epsilon$.==

Answer frame. For the short note, open with the law, write $\sigma=E\epsilon$, name E, G and K with formulas, and state that it holds only up to the proportional limit. For numericals, convert units first, find area, then substitute.

Asked: [7 marks] (Nov 2022, Jun 2022) Find Young's modulus of a brass rod, d = 25 mm, L = 250 mm, load 50 kN, extension 0.3 mm. Rod 200 cm long, d = 3 cm, pull 30 kN, E = 2 x 10^5 N/mm²: find stress, strain, elongation. Asked: [14 marks] (Dec 2023) Write short note on any two: (a) Hooke's law and modulus of elasticity, (b) Micrometer, (c) Newton's law of viscosity, (d) Natural and artificial draught. Pitfall: Convert kN to N and cm to mm before substituting, or E is off by powers of ten.

Hardness and impact testing of materials, BHN

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Definition. Hardness is the resistance of a material to indentation, scratching or abrasion. It is measured by pressing a hard indenter into the surface under a load and measuring the indentation.

Diagram.

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Key points.

  1. Brinell test: a hardened steel or tungsten carbide ball of diameter $D$ (10 mm) is pressed with load $P$ (3000 kg for steel) for 10-30 seconds, and the indentation diameter $d$ is measured with a microscope.
  2. $$BHN=\frac{2P}{\pi D\left(D-\sqrt{D^{2}-d^{2}}\right)}$$ which is load divided by the curved area of the indentation; it suits soft to medium metals.
  3. Rockwell test: a small minor load seats the indenter, then a major load is applied, and the hardness is read directly from the dial as the extra depth of penetration; a diamond cone (scales A, C) or steel ball (scale B) is used. It is quick and direct.
  4. Vickers test: a square-based diamond pyramid (136 degrees) is pressed in, the two diagonals of the impression are averaged, and $HV=1.854P/d^{2}$; it covers all hardness ranges and thin sections.
  5. Impact test: a notched bar is broken by a swinging pendulum (Izod: cantilever; Charpy: simply supported), and the energy absorbed in fracture measures toughness.
  6. Purpose of hardness testing: to check wear resistance, the effect of heat treatment, and material quality without destroying the part.

<mark>Hardness is the resistance of a material to indentation, and the Brinell number is the applied load divided by the curved surface area of the indentation.</mark>

Answer frame. Open with the definition of hardness; draw the Brinell sketch; give the procedure and the BHN formula, then Rockwell (minor and major load, depth) and Vickers (diamond pyramid, diagonals); close with the impact test in one line or the purpose of testing.

Asked: [7 marks] (Nov 2022, Dec 2023, Dec 2024) Explain the methods to measure the hardness of materials. Define hardness and explain the Brinell hardness test in detail. What is the purpose of hardness testing and how is it performed?

Last-minute revision

  • Alloy: homogeneous mixture of two or more metals or a metal and non-metal.
  • Cast iron: 2-4.5% C; types grey, white, malleable, ductile (nodular), compacted graphite.
  • Steel: up to about 1.5% C; eutectoid point 0.8% C at 723 C; cementite 6.67% C.
  • Stainless steel: more than 12% Cr; high-speed steel: tungsten.
  • Hooke's law: $\sigma=E\epsilon$; elongation $\Delta L=PL/(AE)$.
  • Mild steel curve points: proportional limit, elastic limit, upper yield, lower yield, ultimate, breaking.
  • Dec 2023 numerical: 763.94, 509.30, 795.77 MPa and 10%.
  • Brass rod: $E=84.88$ GPa; steel rod: 42.44 N/mm², $2.122\times10^{-4}$, 0.424 mm.
  • Ductility is measured by percentage elongation and reduction in area.
  • BHN = $2P/[\pi D(D-\sqrt{D^2-d^2})]$; Rockwell reads depth; Vickers uses a diamond pyramid.

Memory hooks

  • Points on the curve: "PEYUB", Proportional, Elastic, Yield, Ultimate, Breaking.
  • Brinell = Ball, Rockwell = Reading depth, Vickers = Very hard pyramid.
  • Cast iron types: "Grey White Malleable Ductile", G-W-M-D.
  • Cr keeps steel clean (stainless), W keeps it hot (tool), Ni keeps it tough.

Coverage checklist

  • Classification of engineering material: Nov 2022, Dec 2024, Jun 2025 (classify, alloys, applications).
  • Composition of Cast iron and Carbon steels: Dec 2024 (classification of cast irons).
  • Iron Carbon diagram: no past question; phases and points covered.
  • Alloy steels their applications: Jun 2023 (eight elements), Dec 2024 (necessity, difference).
  • Mechanical properties like strength, hardness, toughness , ductility, brittleness , malleability etc. of materials: Jun 2022, Dec 2023, Jun 2025.
  • Tensile test- Stress-strain diagram of ductile and brittle materials: Jun 2022, Jun 2023, Jun 2025, Dec 2023 numerical, Jun 2023 definitions.
  • Hooks law and modulus of elasticity: Nov 2022, Jun 2022 numericals, Dec 2023 short note.
  • Hardness and Impact testing of materials, BHN etc.: Nov 2022, Dec 2023, Dec 2024.
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